Solve for x 1/7=49^(x-5)*7^(x-9)

$\frac{1}{7} = 49^{x - 5} \cdot 7^{x - 9}$

Rewrite the equation as $49^{x - 5} \cdot 7^{x - 9} = \frac{1}{7}$ .

$49^{x - 5} \cdot 7^{x - 9} = \frac{1}{7}$

Raise $7$ to the power of $1$ .

$49^{x - 5} \cdot 7^{x - 9} = \frac{1}{7^{1}}$

Move $7^{1}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$49^{x - 5} \cdot 7^{x - 9} = 7^{- 1}$

Rewrite $49$ as $7^{2}$ .

${(7^{2})}^{x - 5} \cdot 7^{x - 9} = 7^{- 1}$

Multiply the exponents in ${(7^{2})}^{x - 5}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$7^{2 (x - 5)} \cdot 7^{x - 9} = 7^{- 1}$

Apply the distributive property.

$7^{2 x + 2 \cdot - 5} \cdot 7^{x - 9} = 7^{- 1}$

Multiply $2$ by $- 5$ .

$7^{2 x - 10} \cdot 7^{x - 9} = 7^{- 1}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$7^{2 x - 10 + x - 9} = 7^{- 1}$

Add $2 x$ and $x$ .

$7^{3 x - 10 - 9} = 7^{- 1}$

Subtract $9$ from $- 10$ .

$7^{3 x - 19} = 7^{- 1}$

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$3 x - 19 = - 1$

Solve for $x$ .

Move all terms not containing $x$ to the right side of the equation.

Add $19$ to both sides of the equation.

$3 x = - 1 + 19$

Add $- 1$ and $19$ .

$3 x = 18$

Divide each term in $3 x = 18$ by $3$ and simplify.

Divide each term in $3 x = 18$ by $3$ .

$\frac{3 x}{3} = \frac{18}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{18}{3}$

Divide $x$ by $1$ .

$x = \frac{18}{3}$

Simplify the right side.

Divide $18$ by $3$ .

$x = 6$